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Edit: (Sept 27, 2017) The result I was trying to prove here is false. Not all hyperbolas can be realized as right circular conic sections. Also, the approach described below is not the best way to treat the problem. I posted an arxiv paper with a better approach:

https://arxiv.org/abs/1708.07093

I'm trying to find a concise proof of the following statement:

Given real numbers $a,b,c,d,e,f\in\mathbb{R}$, the real solution of the equation $$ax^2+bxy+cy^2+dx+ey+f=0\tag{1}$$ can be seen as the intersection of the $x,y$-plane with a right circular cone in $x,y,z$-space. (At least in the generic case.)

Here's my attempt at a proof:

First homogenize the equation to obtain $$ax^2+bxy+cy^2+dxz+eyz+fz=0 \tag{2}$$ and then express this as a matrix equation: $$\mathbf{x}^T H \mathbf{x}=\begin{pmatrix} x&y&z\end{pmatrix}\begin{pmatrix}a& b/2 & d/2 \\ b/2 & c & e/2 \\ d/2 & e/2 & f\end{pmatrix}\begin{pmatrix} x\\y\\z\end{pmatrix}=0.\tag{2}$$ Since $H$ is a real symmetric matrix, the Principal Axes Theorem says that there exists a real orthogonal matrix $Q^T=Q^{-1}$ such that $H':=Q^THQ$ is diagonal: $$H':=Q^THQ=\begin{pmatrix} a'&0&0\\ 0&c'&0 \\ 0&0&f' \end{pmatrix}.$$ By making the change of variables $$\begin{pmatrix} x\\y\\z\end{pmatrix}=\mathbf{x}=Q\mathbf{x}'=Q\begin{pmatrix} x'\\y'\\z'\end{pmatrix}$$ we see that equation (2) is equivalent to the following: \begin{align} \mathbf{x}^T H \mathbf{x} &= 0 \\ (Q\mathbf{x}')^T H (Q\mathbf{x}') &= 0 \\ (\mathbf{x}')^T (Q^THQ) \mathbf{x}' &= 0 \\ (\mathbf{x}')^T H' \mathbf{x}'&= 0 \\ a'(x')^2+c'(y')^2+f'(z')^2 &= 0.\tag{3} \end{align}

Now, if it happens that $a',c',f'\in\{\pm 1\}$ then I claim that we are done. Indeed, in this case we identify (3) as the equation of a right circular cone (or a single point, in which case the result is true anyway). Then since (2) and (3) are related by an orthogonal transformation we see that (2) also defines a right circular cone. Finally, we see that (1) defines the intersection of this cone with the plane $z=1$ in $x,y,z$-space. QED

Unfortunately we do not always have $a',c',f'\in\{\pm 1\}$. If $a'c'f'\neq 0$ then equation (3) defines a single point or an elliptic cone in $x',y',z'$-space, which is not quite what I want.

Here is my question:

Is there a way to tweak the argument to force $a',c',f'\in\{\pm 1\}$? My only idea is to perform an initial rotation and translation of equation (1), that is, to make an initial change of variables of the form $$\begin{pmatrix} x'\\y'\end{pmatrix} = P\begin{pmatrix} x\\y\end{pmatrix} +\begin{pmatrix}u\\v\end{pmatrix}$$ with $P^T=P^{-1}$. Hopefully one could choose this $P$ and $(u,v)$ appropriately so the homogenized form of the new equation (1) has eigenvalues in $\{\pm 1\}$ but I don't see how to do it.

Any ideas?

Drew Armstrong
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  • Write each eigenvalue as $sr^2$ where $s = \pm 1$. Then form matrices $S$ and $R$ with these on the diagonal. Now your matrix $Q$ is just $$Q = R^t S R.$$So by changing variables again to $x'' = Rx$, you get $x'' ^t S x'' = 0$ and you've got what you wanted. – John Hughes Apr 09 '17 at 22:07
  • But then the matrix $Q$ is no longer orthogonal, so the geometry is lost when we change variables back to equation (2). – Drew Armstrong Apr 09 '17 at 23:50
  • I looked at your note in arxiv but could not find where it showed there are hyperbolas that cannot be realized by right circular cones. I did see a claim that not all quadric surfaces have right circular tangent cones, but that's a different matter. – David K Jul 29 '18 at 00:04
  • In fact, following the references I gave here, you can construct an ellipse whose major axis is the segment between the hyperbola's foci and whose foci are the vertices of the hyperbola, in a plane perpendicular to the plane of the hyperbola; the cone from any point on that ellipse to the hyperbola is a right circular cone. This works for any non-degenerate hyperbola. – David K Sep 01 '18 at 20:13

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I think I see your problem. You (like me!) read too much into the claim you were trying to prove. You wanted to say that the you could transform things so that you conic would be the intersection of something parallel to the $xy$ plane with a PARTICULAR right circular conic, one whose base lies in the $xy$ plane. But that's not true, for all slices of that conic are circular or single points.

In what you wrote, you probably CAN see how to make the conic be the intersection of your right-circular cone with some plane other than the $xy$ plane --- the drawings for conics you see in any textbook show this.

Well...once you see that, you can rotate space so that your slicing plane is the $xy$ plane. And when you do that, your right-circular cone becomes...a right circular cone ...but one whose base is on some slanted plane in space rather than the $xy$ plane (in general).

John Hughes
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  • Yes, I see that I was looking for something impossible: I wanted to realize an arbitrary conic section as the intersection of the plane $z=1$ in $x,y,z$-space with a right circular cone whose apex is on the plane $z=0$. In general I suppose I will need to look at the planes $z=0$ and $z=t$ for some arbitrary $t$. So here is what the problem boils down to: – Drew Armstrong Apr 10 '17 at 14:27
  • Consider any $3\times 3$ real symmetric matrix $H^T=H$. Find a $2\times 2$ orthogonal matrix $P^T=P^{-1}$, a $2\times 1$ vector $\mathbf{u}$ and a nonzero constant $t$ such that the matrix $$\begin{pmatrix} P & \mathbf{0} \ \mathbf{u}^T & t\end{pmatrix} H \begin{pmatrix} P & \mathbf{u} \ \mathbf{0}^T & t\end{pmatrix}$$ has eigenvalues in the set ${\pm 1}$. I still don't see how to solve it. – Drew Armstrong Apr 10 '17 at 14:36